In a compound,nitrogen atoms (N) make a cubic | vedclass

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(B) Let the number of $N$ atoms in the cubic close-packed (ccp) lattice be $4$. Since the number of tetrahedral voids is twice the number of atoms in

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$M_2N_3$

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(B) Let the number of $N$ atoms in the cubic close-packed (ccp) lattice be $4$. Since the number of tetrahedral voids is twice the number of atoms in the lattice,the number of tetrahedral voids $= 2 	imes 4 = 8$. The metal atoms $(M)$ occupy one-third of these tetrahedral voids,so the number of $M$ atoms $= \frac{1}{3} 	imes 8 = \frac{8}{3}$. The ratio of $M$ atoms to $N$ atoms is $M:N = \frac{8}{3}: 4$. Multiplying both sides by $3$,we get $M:N = 8: 12$,which simplifies to $2: 3$. Therefore,the formula of the compound is $M_2N_3$.

In a compound,nitrogen atoms $(N)$ make a cubic close-packed lattice and metal atoms $(M)$ occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by $M$ and $N$.

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